NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
If ∫ e - x 2 2 d x = f x and the solution of the differential equation d y d x = 1 + x y is y = k e x 2 2 f ( x ) + C e x 2 2 , then the value of k is equal to (where C is the constant of integration)
Correct answer
1
Step-by-step solution
d y d x - x y = 1 Here, I.F. = e - ∫ x d x = e - x 2 2 So, solution is y e - x 2 2 = ∫ e - x 2 2 d x + C y = e x 2 2 f x + C e x 2 2 Hence, k = 1